Hi, I understand the problem of high Peak to Average Power Ratio in OFDM. However, I am not sure how the average power of an OFDM signal is calculated. Suppose you have an N-carrier transmitter where the average power per carrier is p, what would the average power of the OFDM signal be? Thanks.

# OFDM average power

Started by ●September 20, 2005

Reply by ●September 20, 20052005-09-20

Snowball wrote:> Hi, > > I understand the problem of high Peak to Average Power Ratio in OFDM. > However, I am not sure how the average power of an OFDM signal is > calculated. Suppose you have an N-carrier transmitter where the average > power per carrier is p, what would the average power of the OFDM signal be? > > Thanks. > >The 'O' in OFDM stands for "Orthogonal", and the powers of orthogonal signals add linearly. So the total average power would be N * p -- assuming that your assumptions were correct. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com

Reply by ●September 20, 20052005-09-20

Tim Wescott 쓴 글:> The 'O' in OFDM stands for "Orthogonal", and the powers of orthogonal > signals add linearly. So the total average power would be N * p -- > assuming that your assumptions were correct.Plus, the (total) average power as well as the peak power must be estimated in the time-domain. PAPR in OFDM is a problem in the time-domain, i.e., in each time-domain symbol after IFFT at the transmitter. The symbols in the frequency-domain, i.e., before IFFT at the transmitter, are out of interests. The transformed signal can be an infinite power signal although the time-domain symbols are points within the finite constellation. The total average power in the both domains are the same since the full ranked ortho-normal property of IFFT as it was pointed out by Tim. Note that the average power of each basis in two domains can be, obviously, different. The total average power is derived as P_total = tr(E[T'ss'T]) = tr(T'RT) = tr(R) where T and s are the FFT transform matrix and the OFDM symbol vector, respectively, and ()' and tr() are the operator for Hermitian and Trace of the matrix, respectively. In addition, R = E[ss']. - James GOLD

Reply by ●September 21, 20052005-09-21

"Tim Wescott" <tim@seemywebsite.com> wrote in message news:MuGdnUb4oLkHCK3eRVn-3A@web-ster.com...> Snowball wrote: > > > Hi, > > > > I understand the problem of high Peak to Average Power Ratio in OFDM. > > However, I am not sure how the average power of an OFDM signal is > > calculated. Suppose you have an N-carrier transmitter where the average > > power per carrier is p, what would the average power of the OFDM signalbe?> > > > Thanks. > > > > > The 'O' in OFDM stands for "Orthogonal", and the powers of orthogonal > signals add linearly. So the total average power would be N * p -- > assuming that your assumptions were correct. > > -- > > Tim Wescott > Wescott Design Services > http://www.wescottdesign.comThanks for your answers guys. The implication of this would be that for OFDM with N carriers (and the same power and modulation per carrier), you would need N times the power of a single-carrier system to achieve the same BER, correct? Am I missing something obvious?

Reply by ●September 21, 20052005-09-21

Snowball wrote:> "Tim Wescott" <tim@seemywebsite.com> wrote in message > news:MuGdnUb4oLkHCK3eRVn-3A@web-ster.com... > >>Snowball wrote: >> >> >>>Hi, >>> >>>I understand the problem of high Peak to Average Power Ratio in OFDM. >>>However, I am not sure how the average power of an OFDM signal is >>>calculated. Suppose you have an N-carrier transmitter where the average >>>power per carrier is p, what would the average power of the OFDM signal > > be? > >>>Thanks. >>> >>> >> >>The 'O' in OFDM stands for "Orthogonal", and the powers of orthogonal >>signals add linearly. So the total average power would be N * p -- >>assuming that your assumptions were correct. >> >>-- >> >>Tim Wescott >>Wescott Design Services >>http://www.wescottdesign.com > > > Thanks for your answers guys. The implication of this would be that for OFDM > with N carriers (and the same power and modulation per carrier), you would > need N times the power of a single-carrier system to achieve the same BER, > correct? Am I missing something obvious? > >All else being equal, i.e. the same symbol rate. So if you took your OFDM system and removed carriers but kept each modulated carrier the same you should expect the same (realistically slightly better) BER. What you would _not_ be able to do in the environments for which OFDM was designed is use 1 carrier and N times the symbol rate -- OFDM is used to circumvent processes which cause severe intersymbol interference at high symbol rates, which is exactly what you'd get. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com

Reply by ●September 21, 20052005-09-21

Snowball <sdris@softlab.ntua.gr> wrote:>I understand the problem of high Peak to Average Power Ratio in OFDM. >However, I am not sure how the average power of an OFDM signal is >calculated.What is really meant by "average" power in the expression "PAR" is RMS power. Now here's a question for the group. What is the par of a sine wave? (1) 1.0 (2) sqrt(2)/2 Let's hold a vote. Steve

Reply by ●September 22, 20052005-09-22

Steve Pope wrote:> Snowball <sdris@softlab.ntua.gr> wrote: > > >>I understand the problem of high Peak to Average Power Ratio in OFDM. >>However, I am not sure how the average power of an OFDM signal is >>calculated. > > What is really meant by "average" power in the expression "PAR" > is RMS power.I know what RMS voltage and current are. What is RMS power? How is it calculated?> Now here's a question for the group. What is the par of > a sine wave? > > (1) 1.0 > (2) sqrt(2)/2Are these peak values, or RMS? P-P?> Let's hold a vote.Not till we know what we're voting on. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●September 22, 20052005-09-22

"Jerry Avins" <jya@ieee.org> wrote in message news:s56dnc6VBOEpOK_eRVn-ug@rcn.net...> Steve Pope wrote: > > Snowball <sdris@softlab.ntua.gr> wrote: > > > > > >>I understand the problem of high Peak to Average Power Ratio in OFDM. > >>However, I am not sure how the average power of an OFDM signal is > >>calculated. > > > > What is really meant by "average" power in the expression "PAR" > > is RMS power. > > I know what RMS voltage and current are. What is RMS power? How is it > calculated? > > > Now here's a question for the group. What is the par of > > a sine wave? > > > > (1) 1.0 > > (2) sqrt(2)/2 > > Are these peak values, or RMS? P-P? > > > Let's hold a vote. > > Not till we know what we're voting on. > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������I agree with Jerry :) What is the definition of rms power?

Reply by ●September 22, 20052005-09-22

Jerry Avins <jya@ieee.org> wrote:>Steve Pope wrote:>> What is really meant by "average" power in the expression "PAR" >> is RMS power.>I know what RMS voltage and current are. What is RMS power? How is it >calculated?I'm being sloppy, I guess. I meant power computed by an RMS method. If it's a voltage signal, compute the RMS voltage V, then power is V^2/R. But often in signal processing, one has an amplitude signal that is not dimensionalized as voltage, current, or anything else. In a sense it's meaningless to speak of the power of such a signal until you have more units involved, but people often do.>> Now here's a question for the group. What is the par of >> a sine wave? >> >> (1) 1.0 >> (2) sqrt(2)/2>Are these peak values, or RMS? P-P?The way I see it, if we can't agree on the PAR of a sine wave, how can we agree on the PAR of a more complex signal such as OFDM? Steve

Reply by ●September 23, 20052005-09-23

Steve Pope wrote:> Jerry Avins <jya@ieee.org> wrote: > > >>Steve Pope wrote: > > >>>What is really meant by "average" power in the expression "PAR" >>>is RMS power. > > >>I know what RMS voltage and current are. What is RMS power? How is it >>calculated? > > > I'm being sloppy, I guess. I meant power computed by an RMS > method. If it's a voltage signal, compute the RMS voltage V, > then power is V^2/R.Only for sinusoids. To make that work for other waveshapes, one must define V = sqrt(R*P), which is tautological.> But often in signal processing, one has an amplitude signal that > is not dimensionalized as voltage, current, or anything else. > In a sense it's meaningless to speak of the power of such > a signal until you have more units involved, but people often do.Yes. Many computations and concepts are simplifies by assuming a voltage -- as is common -- or current. Unity power factor is also assumed. We need to bear in mind the important distinction that these are simplifying assumptions, and nor reality.>>>Now here's a question for the group. What is the par of >>>a sine wave? >>> >>>(1) 1.0 >>>(2) sqrt(2)/2 > > >>Are these peak values, or RMS? P-P? > > > The way I see it, if we can't agree on the PAR of a sine > wave, how can we agree on the PAR of a more complex signal > such as OFDM?The PAR of a sine is well defined. I asked if 1.0 and sqrt(2) represented RMS, peak, or P-P vales. Without an answer to that, there are no numerical answers to the numerators and denominators in your question. Strictly, in the equation A = sin(wt), A is the amplitude -- a constant independent of time -- and the peak instantaneous value. 2*A is the P-P value, and A/sqrt(2) is the RMS value. All these are independent of power and impedance. With the usual simplifying assumptions (Z = R = 1), average power (what I think you meant by "RMS power") is A^2/2 and peak power is A^2, giving a PAR (for a sine of any phase and amplitude) of 2. It's too late for agreement; that definition is established. Jerry P.S. In real circuits, power factor (the cosine of the angle between voltage and current) matters. As PF goes to zero, PAR goes to infinity. -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������